Example 1.5.3 | 例1.5.3

📚 Example 1.5.3 | 例1.5.3

This worked example, taken from the AQA A-Level Mathematics specification, demonstrates how to write a quadratic expression in completed-square form and how to use that form to analyse the graph of the function. We will follow Example 1.5.3 step by step, making sure every algebraic manipulation is clear and justified.

这个示例来自 AQA A-Level 数学大纲，演示如何将二次表达式写成完全平方形式，并利用该形式分析函数图像。我们将逐步跟随例 1.5.3，确保每一步代数变形清晰且有理有据。

1. The Problem | 问题描述

The task is to rewrite the quadratic expression y = 3x² − 12x + 7 as a(x + p)² + q. Here a, p and q are constants to be determined. After completing the square, we must state the minimum value of y and the coordinates of the vertex.

任务是将二次表达式 y = 3x² − 12x + 7 改写为 a(x + p)² + q 的形式。其中 a、p、q 为待定常数。完成配方后，我们必须说明 y 的最小值和顶点坐标。

2. Why Complete the Square? | 为什么配方？

Completing the square transforms a quadratic expression into a form that clearly shows the horizontal and vertical translations of the basic curve y = x². The vertex is then easy to read: ( −p , q ), and the minimum (or maximum) value is q when a > 0 (or a < 0).

配方法将二次表达式转换为一种能清晰显示基本曲线 y = x² 的水平和垂直平移的形式。顶点因此容易读出：( −p , q )，并且当 a > 0 时最小值为 q（当 a < 0 时最大值为 q）。

3. Factor Out the Leading Coefficient | 提取二次项系数

In the expression 3x² − 12x + 7, the coefficient of x² is 3. We first factor 3 out of the first two terms, leaving the constant outside:

在表达式 3x² − 12x + 7 中，x² 的系数为 3。我们先把 3 从第一项和第二项中提出来，常数项留在外面：

y = 3(x² − 4x) + 7

This step respects the order of operations: the 3 multiplies only the x² and −4x terms, not the constant 7.

这一步遵循运算顺序：3 只乘以 x² 和 −4x 项，不乘常数 7。

4. Complete the Square Inside the Bracket | 在括号内配方

Now we focus on x² − 4x. To complete the square, we take half of the coefficient of x, which is −4/2 = −2, and square it to get (−2)² = 4. Hence:

现在关注 x² − 4x。要配方，我们取 x 的系数的一半，即 −4/2 = −2，然后平方得到 (−2)² = 4。因此：

x² − 4x = (x − 2)² − 4

Check: expanding (x − 2)² gives x² − 4x + 4, so subtracting 4 returns x² − 4x.

检查：展开 (x − 2)² 得到 x² − 4x + 4，减去 4 就回到 x² − 4x。

5. Substitute Back and Simplify | 代回并化简

Replace x² − 4x in the factored expression:

在提取后的表达式中替换 x² − 4x：

y = 3[(x − 2)² − 4] + 7

Then distribute the 3 and combine constant terms:

然后展开 3 并合并常数项：

y = 3(x − 2)² − 12 + 7 = 3(x − 2)² − 5

Thus a = 3, p = −2, q = −5, because the form a(x + p)² + q uses a plus sign before p.

因此 a = 3，p = −2，q = −5，因为形式 a(x + p)² + q 中 p 前是加号，所以此处相当于加上 −2。

6. Identify the Vertex | 确定顶点

In the completed-square form y = a(x + p)² + q, the vertex is located at (−p, q). In our expression y = 3(x − 2)² − 5, we have p = −2, so −p = 2. Therefore the vertex is (2, −5).

在完全平方形式 y = a(x + p)² + q 中，顶点位于 (−p, q)。在我们的表达式 y = 3(x − 2)² − 5 中，p = −2，所以 −p = 2。因此顶点为 (2, −5)。

Because a = 3 > 0, the parabola opens upwards, and the vertex represents the minimum point.

因为 a = 3 > 0，抛物线向上开口，顶点代表最小值点。

7. State the Minimum Value | 写出最小值

At the vertex, x = 2 and y = q = −5. Because the curve opens upwards, this is the global minimum value of the function. Thus the minimum value of y is −5.

在顶点处，x = 2，y = q = −5。由于曲线向上开口，这是函数的全局最小值。因此 y 的最小值是 −5。

This can also be reasoned from the square term: 3(x − 2)² is always non-negative, reaching 0 only at x = 2, so y is never less than −5.

这也可以从平方项推理：3(x − 2)² 始终非负，仅在 x = 2 时取 0，所以 y 永远不会小于 −5。

8. Axis of Symmetry | 对称轴

The vertical line through the vertex is the axis of symmetry. Its equation is x = 2. This means that the graph is a reflection about this line; inputs equidistant from x = 2 produce the same y-value.

经过顶点的竖直线是对称轴。它的方程是 x = 2。这意味着图像关于该线对称；距离 x = 2 等距的输入产生相同的 y 值。

For example, x = 1 and x = 3 both give y = 3(1 − 2)² − 5 = −2, confirming the symmetry.

例如，x = 1 和 x = 3 都给出 y = 3(1 − 2)² − 5 = −2，验证了对称性。

9. Check by Expanding | 展开验证

To verify the completed square, we can expand 3(x − 2)² − 5. First, 3(x² − 4x + 4) = 3x² − 12x + 12. Then subtract 5 gives 3x² − 12x + 7, which matches the original expression. This confirms the algebra.

为验证配方，我们可以展开 3(x − 2)² − 5。首先，3(x² − 4x + 4) = 3x² − 12x + 12。然后减去 5 得到 3x² − 12x + 7，与原表达式一致。这确认了代数正确。

Such an expansion check is a valuable habit in an exam: it catches sign errors and arithmetic mistakes without taking extra time.

这种展开检验在考试中是非常好的习惯：它能发现符号错误和算术失误，而且不花额外时间。

10. Sketching the Graph | 绘制图像

Using the vertex (2, −5), the axis of symmetry x = 2, and the y-intercept (found by setting x = 0 in the original equation: y = 7), we can make a rough sketch. The curve is a parabola with a vertical stretch factor of 3, so it is narrower than y = x².

利用顶点 (2, −5)、对称轴 x = 2 和 y 截距（在原方程中令 x = 0：y = 7），我们可以画一个粗略的草图。曲线是抛物线，垂直拉伸因子为 3，因此比 y = x² 更窄。

A careful sketch should also show that the graph crosses the x-axis twice, because the minimum point is below the x-axis.

仔细的草图还应显示图像与 x 轴相交两次，因为最小值点在 x 轴下方。

11. Connections to the Discriminant | 与判别式的联系

The discriminant of the original quadratic is Δ = (−12)² − 4 × 3 × 7 = 144 − 84 = 60, which is positive. This tells us the graph has two real roots. Notice that the minimum value −5 is below the x-axis, so indeed the curve crosses the x-axis in two places.

原二次方程的判别式为 Δ = (−12)² − 4 × 3 × 7 = 144 − 84 = 60，为正。这告诉我们图像有两个实根。注意最小值 −5 在 x 轴下方，因此曲线确实在 x 轴上有两个交点。

The completed-square form can also be used directly to solve y = 0: 3(x − 2)² = 5, so x = 2 ± √(5/3). This gives the exact roots.

完全平方形式也可以直接用来解 y = 0：3(x − 2)² = 5，所以 x = 2 ± √(5/3)。这表明精确根。

12. Summary | 总结

We have shown how to complete the square for the quadratic expression 3x² − 12x + 7. The completed-square form is 3(x − 2)² − 5. From this we read the vertex (2, −5), the minimum value −5, and the axis of symmetry x = 2. The same method applies to any quadratic with non-zero leading coefficient.

我们已经演示了如何对二次表达式 3x² − 12x + 7 配方。完全平方形式为 3(x − 2)² − 5。由此我们读出顶点 (2, −5)、最小值 −5 和对称轴 x = 2。同样的方法适用于任何首项系数非零的二次表达式。

Remember that in the form a(x + p)² + q, the vertex is always (−p, q). If a > 0, the turning point is a minimum; if a < 0, it is a maximum.

请记住在形式 a(x + p)² + q 中，顶点始终是 (−p, q)。如果 a > 0，转向点是最小值；如果 a < 0，则是最大值。

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Example 1.5.3 | 例1.5.3